We have the equation of motion [tex]s = ut + \frac{1}{2} at^2[/tex], where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.
Here s = 427 m, u = 0 m/s, a = 9.81 [tex]m/s^2[/tex]
Substituting
[tex]427 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 = 427\\ \\ t =9.335 seconds[/tex]
Now we have v = u+at, where v is the final velocity
Here u = 0 m/s, a= 9.81 [tex]m/s^2[/tex] and t = 9.335 seconds
Substituting
v = 0+9.8*9.335 = 91.483 m/s
The speed with which the penny strikes the ground = 91.483 m/s.
The speed is 91.52 m/sec at which the penny strikes the ground.
Given :
Height of the building is 427m.
Solution :
We know that displacement is,
[tex]\rm s= ut+\dfrac{1}{2}at^2[/tex] ----- (1)
Where,
initial velocity, u = 0
t is time,
Gravitational acceleration, [tex]\rm a =9.81\;m/sec^2[/tex]
Now putting the values of a, s and u in equation (1),
[tex]\rm 427 = 0\times t + \dfrac{1}{2}\times 9.81\times t^2[/tex]
[tex]t = \sqrt{\dfrac{427\times2}{9.81}}[/tex]
[tex]\rm t =9.33 \;sec[/tex]
Now we also know that,
[tex]\rm v = u + at[/tex] ----- (2)
Where,
[tex]\rm a = 9.81\;m/sec[/tex]
t = 9.33 sec
u = 0 m/sec
putting the values of a, u and t in equation (2),
[tex]\rm v = 0+9.81\times 9.33[/tex]
[tex]\rm v =91.52\;m/sec[/tex]
The speed is 91.52 m/sec at which the penny strikes the ground.
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https://brainly.com/question/2987794?referrer=searchResults
